Problem: Evaluate the definite integral. $\int^1_{-1}\left(\dfrac{27}{x^4}-3\right)\,dx = $
Explanation: First, use the power rule: $\begin{aligned}\int^1_{-1}\left(\dfrac{27}{x^4}-3\right)\,dx~&=~\int^1_{-1}\left({27}{x^{-4}}-3\right)\,dx \\&=(-9x^{-3}-3x)\Bigg|^1_{{-1}}\end{aligned}$ Second, plug in the limits of integration: $[-9\cdot1^{-3}-3\cdot1]-[-9\cdot({-1})^{-3}-3\cdot({-1})] = -12-12 = -24$. The answer: $\int^1_{-1}\left(\dfrac{27}{x^4}-3\right)\,dx = -24$